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Question Number 108825 by ajfour last updated on 19/Aug/20

$$\underset{n=1}{\sum ^{10}}(i^n+i^{n+1})=?$$

Commented by ajfour last updated on 19/Aug/20

$$thankseveryonewhoconfirmed.$$

Answered by john santu last updated on 19/Aug/20

$$\frac{⊸JS⊸}{\mathrm{♡}}$$$$\underset{n=1}{\sum ^{10}}{i}^n(1+i)=(1+i)\underset{n=1}{\sum ^{10}}{i}^n=(1+i)(i-1)=-1-1=-2$$$$\underset{k=1}{\sum ^{10}}{i}^n=i-1-i+1+i-1-i+1+i-1$$$$=i-1(\ast )$$$$$$

Commented by ajfour last updated on 19/Aug/20

$$\underset{k=1}{\sum ^{10}}{i}^n=i-1-i+1+i-1-i+1+i-1$$$$=i-1$$$$so{\sum _{n=1}}^{10}{i}^n(1+i)=(i+1)(i-1)$$$$=-1-1=-2.$$$${isn}^{\prime }tthisright?$$

Commented by bemath last updated on 19/Aug/20

$$yes$$

Commented by john santu last updated on 19/Aug/20

$$yes$$

Answered by Dwaipayan Shikari last updated on 19/Aug/20

$$i\frac{i^{10}-1}{i-1}+i^2\frac{i^{10}-1}{i-1}=i(i+1)-(i+1)=-1-1=-2$$

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